Numbers: Rational numbers
Ordering of rational numbers
In order to determine which of two rational numbers is the bigger, the next notion is important.
Having common denominators
Two fractions are said to have a common denominator if their denominators are equal.
Putting over a common denominator
Two fractions can always be rewritten so they have a common denominator by multiplying the numerator and denominator of each fraction by the quotient of the least common multiple of the two denominators by the denominator of the other fraction.
In order to put two fractions over a common denominator you should choose a common denominator. This can for example be the product of the denominators. The most efficient method is choosing the least common multiple (lcm) of the original denominators.
In order to put #\frac{1}{6}# and #\frac{1}{3}# over a common denominator we choose #{\rm lcm}(6,3)=6# as the common denominator. The first fraction already has this denominator. The second fraction the numerator and denominator need to be multiplied by #2#: #\frac{1}{3}=\frac{2}{6}#.
In order to put #\frac{1}{6}# and #\frac{3}{4}# over a common denominator we choose #{\rm lcm}(6,4)=12# as the common denominator. The first fraction can be rewritten by multiplying numerator and denominator by #2#: #\frac{1}{6}=\frac{2}{6\times 2}=\frac{2}{12}#. The second fraction can be rewritten as #\frac{3}{4}=\frac{3\times3}{4\times3}=\frac{9}{12}#.
In the case of more than two fractions, it is possible to consider the first two fractions, next putting this result and the third one over a common denominator, and so on. You can also do it in one go by using the least common multiple of all the denominators.
Comparison of two rational numbers
Let #a#, #b#, #m#, and #n# be integers with #m\gt0# and #n\gt0#.\[ \begin{array}{rcl}\dfrac{a}{m}\lt \dfrac{b}{n}&\text{ if and only if }& a\cdot n \lt b\cdot m\tiny. \end{array}\]In particular, for two fractions with a common denominator, \[ \begin{array}{rcl}\dfrac{a}{n}\lt \dfrac{b}{n}&\text{ if and only if }& a\lt b\tiny. \end{array}\]
In order to derive the statements, we will use the following rule, which is valid for any three numbers #x#, #y#, and #z# with #z\gt0#:
If #z\gt0#, then #x\gt y# if and only if #x\cdot z\gt y\cdot z#.
By this rule, the inequality #\frac{a}{m}\lt \frac{b}{n}# is equivalent to #\frac{a}{m}\cdot m\cdot n \lt \frac{b}{n}\cdot m\cdot n# (take #z=m\cdot n#). By the The product fractions, the left hand side is equal to #a\cdot n# and the right hand side is equal to # b\cdot n#, so the inequality can be rewritten as #{a\cdot n\lt b\cdot m}#.
For proving the statement regarding the case of common denominator #n#, it suffices to multiply both fractions with #n# rather than #m\cdot n#.
For general denominators the unequal sign switches as many times are negative denominators. For example, if #m\gt0# and #n\lt0#, then \[ \begin{array}{rcl}\frac{a}{m}\lt \frac{b}{n}&\text{ if and only if }& a\cdot n\gt b\cdot m\tiny. \end {array}\]
The fastest method does this by
- first determining the least common multiple of the two denominators: \(\mathrm{lcm}(10,4)=\frac{10\times 4}{\mathrm{gcd}(10,4)}=\frac{40}{2}=20\),
- next adjusting the numerators accordingly: \(\frac{2}{10}=\frac{2\times 2}{10\times 2}=\frac{4}{20}\) and \(\frac{1}{4}=\frac{1\times 5}{4\times 5}=\frac{5}{20}\).
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